I am studying for the technician exam and I have come across the section on carrier waves, basebands, and various kinds of modulation.

It makes sense to me that you create a signal by combining the baseband, which carries data, with the carrier signal (a sine wave of some kind) before transmitting.

My question is - why? What would stop us from just transmitting the raw data (voice or whatever) as it is? My guess is that "voice" frequencies for example are in the range of 300hz - 3000hz and we need to "boost" them up to the frequency we are on (for example 167.000 MHz). However this is just a guess.

  • $\begingroup$ Even if you could find some way to send voice as "voice" range frequencies, how would you prevent one signal from interfering with every other signal? $\endgroup$ Jan 16, 2018 at 4:29
  • $\begingroup$ @GregHewgill I'm not sure, I guess I don't understand how using a sine wave prevents interference - unless all the sign waves are just slightly different? $\endgroup$
    – user11031
    Jan 16, 2018 at 5:23

3 Answers 3


The simplest answer is that the modulating signal, while it could be electromagnetic, will not have the desired propagation characteristics to carry out effective communications. By mixing the modulating signal with the carrier, the resulting signal has a much higher frequency that results in the desired propagation effects.

Hams often find that they cannot make their desired contacts on a given band (short frequency range) so they switch bands in the hope of making a contact. The modulating signal stays the same but the carrier frequency is dramatically changed as they switch bands.

Having the ability to change the carrier frequency also helps to avoid interfering signals or talk with another ham by tuning to their transmitting frequency.

  • 1
    $\begingroup$ I would like to add to this that a half wave antenna at 1,500Hz would be about 100Km long. $\endgroup$
    – Lance
    Jan 16, 2018 at 15:29
  • 1
    $\begingroup$ @Lance, yeah but you can talk to submarines 9,000 miles away :) $\endgroup$
    – Wossname
    Jan 19, 2018 at 19:13
  • $\begingroup$ Yeah, then there's that! $\endgroup$
    – Lance
    Jan 19, 2018 at 22:56
  • $\begingroup$ But not at audio frequencies - you still need a carrier. $\endgroup$
    – Glenn W9IQ
    Jan 19, 2018 at 23:05

This really depends on what you consider to be a "carrier wave".

If you consider it a sine wave at RF, then it may or may not be present. For example, an analog AM or FM station transmitting silence is just a pure tone: the carrier.

But silence in single-sideband (SSB) doesn't have a transmitted carrier: in fact if there's no audio doing into the microphone, the transmitter doesn't transmit anything. This is because is simply the baseband signal (voice into the microphone, for example) shifted up in frequency. If the transmitter is tuned to 10,000,000 Hz, and you whistle in the microphone a 500 Hz tone, the transmitter shifts that up by 10,000,000 Hz and transmits a tone at 10,000,500 Hz. In a sense, this is "transmitting the raw data".

However, 10,000,000 Hz is still the carrier frequency. Although it's not part of the transmission, the transmitter somehow needs to convert the baseband frequencies (your voice) up to RF. It does this by multiplying the baseband signal with a 10,000,000 Hz oscillator in a mixer. This alone is effectively an AM transmitter, so to make an SSB transmitter requires additional complexity to remove the second sideband from the transmission.


Let's start with sound. Sound itself is a series of positive and negative waves meaning the sound pressure becomes higher and lower than the surrounding air. That's how the sound propagates. AM transmissions are similar in that the signal starts as the carrier and then the modulation makes the signal strength more or less. What eventually comes out of the transmitter (we're still talking AM here) is the carrier wave plus the heterodyne products [q.v.], the carrier wave frequencies plus (and minus) the audio frequencies. So that's why we need a carrier wave.

So what happens if you made a fancy filter and removed the carrier (and what the heck, how about the "Minus" heterodyne product while we're at it since it's just a mirror image of the "Plus" heterodyne), what do you have? Effectively, you have your audio frequencies shifted up to RF and nothing else. That's called Single Side-Band (SSB.) And just to complete the thought, in order to receive SSB so that you can understand it, you need to reintroduce the carrier wave in the receiver itself or else all you'll hear is something like a duck quacking.

And one more thing, FM is completely different, this analogy doesn't work.

  • 2
    $\begingroup$ I really don't like the wording of this answer: "sound is a series of […] waves". No, it's not. It's a longitudinal pressure wave, not a series of waves. "Than the surrounding air": No. That's not true: Sound waves don't have "surrounding air". It's their medium. The absolute pressure doesn't matter for sound. "What […] comes out of the transmitter […] is the carrier wave plus the heterodyne products": not generally true. AM in its pure form really just is the product of carrier frequency oscillation and audio, no plus there (though technically, things can be different). $\endgroup$ Jan 16, 2018 at 20:20
  • 1
    $\begingroup$ The second paragraph addresses things that OP didn't ask. And the third even moreso. Not sure why OP accepted this as answer. $\endgroup$ Jan 16, 2018 at 20:22
  • $\begingroup$ The second paragraph answers his comment about "boost" frequencies, which I admit, is not a question, but that's why that paragraph is there. The third is to prevent nitpickers from saying "But FM is different." $\endgroup$
    – Duston
    Jan 16, 2018 at 21:25
  • $\begingroup$ It was the only answer for hours, I figured I wasnt going to get another one and the explanation seemed fine to this noob. $\endgroup$
    – user11031
    Jan 17, 2018 at 0:14
  • $\begingroup$ @rec I'm pretty sure that you can unmark it by clicking the green checkmark again, if that's what you want to do. $\endgroup$ Jan 17, 2018 at 3:28

You must log in to answer this question.